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Data for a certain biology experiment are given in the table [#permalink]
29 Sep 2009, 19:29

3

This post received KUDOS

00:00

A

B

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D

E

Difficulty:

55% (hard)

Question Stats:

60% (01:52) correct
40% (01:45) wrong based on 117 sessions

AMOUNT OF BACTERIA PRESENT Time Amount 1:00 P.M. 10.0 grams 4:00 P.M. x grams 7:00 P.M. 14.4 grams

Data for a certain biology experiment are given in the table above. If the amount of bacteria present increased by the same fraction during each of the two 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.?

Re: AMOUNT OF BACTERIA PRESENT [#permalink]
29 Sep 2009, 20:46

2

This post received KUDOS

tejal777 wrote:

I am sorry its not clear..could you please give a detailed expalnation? got my mistake though..its says same fraction..

Sure. As we know that the bacterias are increasing at a constant Fraction, we can write the following equation:

10 + 10*F = X, where F = Fraction by which bacterias increases, and X = amount of bacterias after 3 hrs. After 6 hrs, number of bacterias will be: X+X*F = 14.4 Substituting value of X from above equation, X+X*(X-10)/10 = 14.4 X^2=144 X=12

I hope it is clear now _________________________ Consider KUDOS for good posts

Re: AMOUNT OF BACTERIA PRESENT [#permalink]
29 Sep 2009, 20:00

1

This post received KUDOS

tejal777 wrote:

AMOUNT OF BACTERIA PRESENT Time Amount 1:00 P.M. 10.0 grams 4:00 P.M. x grams 7:00 P.M. 14.4 grams

Data for a certain biology experiment are given in the table above. If the amount of bacteria present increased by the same fraction during each of the two 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.?

A. 12.0 B. 12.1 C. 12.2 D. 12.3 E. 12.4

Please point out my flaw: We know, in 6 hrs. bacteria increased 14.4-10=4.4 So,since it is increasing by a constant, in 3 hrs>> 4.4 x 3/6 = 2.2

Re: AMOUNT OF BACTERIA PRESENT [#permalink]
20 Jul 2011, 07:14

I was using a rather stupid way..and I got B. Can anyone pls tell what got wrong with my calculation? thanks ..

Step 1: 14.4 - 10.00 = 4.4 (the total increase of amount) Step 2: since there are 6 hours, I took 4.4/6 = approx. 0.73 (rounding up to 0.7)

So at 2pm -----> 10.0 + 0.7 = 10.7 3pm -----> 10.7 + 0.7 = 11.4 4pm -----> 11.4 + 0.7 = 12.1 (Here you go answer B) < But it isn't not right answer on OA WHY??

Re: AMOUNT OF BACTERIA PRESENT [#permalink]
20 Jul 2011, 08:47

miweekend wrote:

I was using a rather stupid way..and I got B. Can anyone pls tell what got wrong with my calculation? thanks ..

Step 1: 14.4 - 10.00 = 4.4 (the total increase of amount) Step 2: since there are 6 hours, I took 4.4/6 = approx. 0.73 (rounding up to 0.7)

So at 2pm -----> 10.0 + 0.7 = 10.7 3pm -----> 10.7 + 0.7 = 11.4 4pm -----> 11.4 + 0.7 = 12.1 (Here you go answer B) < But it isn't not right answer on OA WHY??

Can anyone pls enlighten? thanks...

The question says that bacteria increased by same fraction, not by same amount in 2 intervals of 3 hours.

Let X represent the amount of bacteria present at 4:00 PM. Since the fractional increase must remain constant from 1 to 4pm as it is from 4pm to 7pm:

Fractional increase from 1 PM to 4 PM = X / 10.0 Fractional increase from 4 PM to 7 PM = 14.4 / X

X \ 10 = 14.4 \ X

X^2 = 14.4 * 10

X^2 = 144

X = 12

Therefore, the correct answer is A: 12.0. _________________

Re: AMOUNT OF BACTERIA PRESENT [#permalink]
20 Jul 2011, 17:58

varunmaheshwari wrote:

miweekend wrote:

I was using a rather stupid way..and I got B. Can anyone pls tell what got wrong with my calculation? thanks ..

Step 1: 14.4 - 10.00 = 4.4 (the total increase of amount) Step 2: since there are 6 hours, I took 4.4/6 = approx. 0.73 (rounding up to 0.7)

So at 2pm -----> 10.0 + 0.7 = 10.7 3pm -----> 10.7 + 0.7 = 11.4 4pm -----> 11.4 + 0.7 = 12.1 (Here you go answer B) < But it isn't not right answer on OA WHY??

Can anyone pls enlighten? thanks...

The question says that bacteria increased by same fraction, not by same amount in 2 intervals of 3 hours.

Let X represent the amount of bacteria present at 4:00 PM. Since the fractional increase must remain constant from 1 to 4pm as it is from 4pm to 7pm:

Fractional increase from 1 PM to 4 PM = X / 10.0 Fractional increase from 4 PM to 7 PM = 14.4 / X

X \ 10 = 14.4 \ X

X^2 = 14.4 * 10

X^2 = 144

X = 12

Therefore, the correct answer is A: 12.0.

thank you very much....i don't want to sound v. silly but still very confused about the question asked.

......If the amount of bacteria present increased by the same fraction during each of the two 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.?i ....

If 10.0 grams is the denominator so why would 14.4 be the numerator? since 10.0 is at the bottom, 14.4 should be remaining at the bottom too ya?

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